# Find T with Subspace S as Kernel of T

• Bacle2
In summary, the conversation discusses the difficulties in tutoring linear algebra and constructing a map T with a kernel of S, as well as extending this to the infinite-dimensional case. It is mentioned that there is an area of mathematics known as operator algebras that studies non-linear algebra and theorems show how to define F(O) for certain constraints. The conversation also discusses the use of topologies and convergence in studying continuity, as well as the importance of imposing restrictions on infinite-dimensional vectors.
Bacle2
Hi, All:

I have been tutoring linear algebra, and my student does not seem to be able

to understand a solution I proposed ( of course, I may be wrong, and/or explaining

poorly). I'm hoping someone can suggest a better explanation and/or a different solution

to this problem:

We have two vector spaces V,V' , over the same F, and we have a subspace S of V.

The goal is to construct a map T , whose kernel is precisely S. The dimensions of

S,V,W respectively work well re Rank-Nullity, i.e., DimV-DimS=DimV'. My goal is to

declare T to be zero in B_S , and then set up a bijection between the remaining basis

vectors in B_V , and the basis vectors in B_V'.

So, I propossed that:

i)We choose a basis B_S :={e_1,...,e_s} for S, extend to a basis B_V:= {e_1,e_2,...,e_b}

for V. Let B_W:={e_1',e_2',...,e_w'} ; s:=|B_S|, and so-on.

ii)Declare/define T(B_S)==0 , i.e., for each basis vector e_s in B_S, we define

T(e_s)=0

iii) Now, we set up a bijection between the basis vectors in B_V\B_S , and those in

B_W. This bijection, gives rise to an isomorphism (extending by linearity) between

Span(B_V-B_S) , and V' , so we have:

1)T(e_s)==0 , for e_s in B_S

2)T(e_s+i):=e_i'

3)T(ce_s+i+de_s+j):=cT(e_s+i)+dT(e_s+j)

Now, I'm trying to extend this to the infinite-dimensional case, but my student has

only a beginners' knowledge of set theory.

Any Suggestions?

P.S: She also asked me a sort-of-strange question: is there such a thing as "non-linear

Hey Bacle2.

For "non-linear algebra", there is an area of mathematics that is known as operator algebras and there are theorems that show how to given an operator O with certain constraints, you can find the definition for F(O). For example you can find the operator corresponding to SQRT(O) or e^(O) where O is a linear operator.

Also with regard to your 'extend to the infinite-dimensional case', things get really tricky because you can't just apply your finite-dimensional theorems to infinite-dimensional problems in general.

The reason has to do with the nature of infinity. Just like we need convergence for power series, we also need these kinds of properties for associates spaces, basis, and other things in infinite dimensional contexts. It's an absolute pain but it's got to do the fact that infinity just changes the game completely.

If you have a basis matrix for basis A and one for basis B then you derive the basis transformations using the fact that if z is the vector in the universal basis (orthonormal cartesian with a nxn identity matrix as basis) then we have

Uz = a
Vz = b where a is in basis A and b is in basis B with U being basis transformation for basis A and V is same for basis B. You can use the above to find information about going from one basis to another in terms of the actual matrix transformations and if we assume that U and V are proper bases then all the stuff you've talked about before (Rank nullity and so on) has to hold.

If you are talking about non-linear basis transformations you need to use the tensor identities for this, but I'm assuming that you are only referring to linear systems which means the above will hold.

In terms of the infinite-dimensional case, you will need to show specifically that properties of the norm will converge and this means that you will need to consider convergence for an 'infinite by infinite' kind of matrix. If you show that the appropriate convergence properties hold then you can show the appropriate identities for the infinite dimensional cases.

Also is something is a bijection, then the dimension of the basis should not change. If you are mapping say a vector x to y from space A to B and the space is linear in the way that we can represent things by a transformation matrix, then A and B will the same dimension and the basis have to also be the same dimension in terms of the operators involved.

Thanks, Chiro; re non-linear algebra, now that you mention your examples, I think

the Borel Functional Calculus, where one designs a calculus for operators, where one

treats operators as numbers, i.e., we can define square roots, logs, etc. on operators.

Is that what you were referring to?

Re the infinite-dimensional case, you are right; I had not really considered the issue

of rank-nullity. Otherwise, the issue of bijection goes thru, but you're right, there are

Bacle2 said:
Thanks, Chiro; re non-linear algebra, now that you mention your examples, I think

the Borel Functional Calculus, where one designs a calculus for operators, where one

treats operators as numbers, i.e., we can define square roots, logs, etc. on operators.

Is that what you were referring to?

Re the infinite-dimensional case, you are right; I had not really considered the issue

of rank-nullity. Otherwise, the issue of bijection goes thru, but you're right, there are

Yeah the functional analysis is where this kind of thing is considered, but there is a specific area of operator algebras which pays a lot of attention and studies these kinds of things specifically.

What happens is that topologies are brought in since they are one of the more general ways to study continuity and which end up being used to study convergence since if something doesn't converge, it doesn't hold any water no matter what structure, function, whatever we are talking about.

Also another thing is that for infinite-dimensional vectors, you have to impose restrictions on these in addition to the basis because we usually consider these things in a space (like little l^1 or l^2) and for something to be in this space, it has to satisfy a property with respect to some kind of norm in the way that the norm identity of the actual vector converges.

If you want to get a better idea of this, take a look at Hilbert spaces: this is kind of the motivation for all this infinite-dimensional stuff and one thing that is studied intensely are infinite-dimensional systems that appear in quantum mechanics: this is what Von Neumann was working on and its not surprising that he was actually working under Hilbert directly among others which no doubt, helped him build the first rigorous mathematical foundation for QM.

Also before I forget, you should take into account the property of linearity especially for operators.

The main linearity conditions are basically f(aX+bY) = af(X) + bf(Y) and if you find any case where this is violated in terms of an operator, then you by definition have a non-linear operator.

Thanks, Chiro, I figured out a way.

## 1. What is a subspace?

A subspace is a subset of a vector space that satisfies three properties: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication. In other words, if you add two vectors in the subspace, the resulting vector will also be in the subspace. Similarly, if you multiply a vector in the subspace by a scalar, the resulting vector will also be in the subspace.

## 2. What is a kernel of a linear transformation?

The kernel of a linear transformation is the set of all vectors in the domain that get mapped to the zero vector in the range. In other words, it is the set of all input vectors that will result in an output of zero. The kernel can also be thought of as the set of all vectors that are "ignored" or "collapsed" by the linear transformation.

## 3. How do you find T with Subspace S as the kernel?

To find T, we need to determine the linear transformation itself. One way to do this is to consider the basis vectors of Subspace S and how they are transformed by T. Since Subspace S is the kernel of T, we know that all the basis vectors will be mapped to the zero vector. From there, we can construct the matrix representation of T using the standard basis vectors of the vector space.

## 4. What is the significance of the kernel being a subspace?

The fact that the kernel is a subspace is significant because it allows us to use the properties of subspaces to better understand and manipulate the linear transformation. For example, we can use the properties of subspaces to determine the dimension of the kernel and to find a basis for the kernel.

## 5. Can a linear transformation have more than one subspace as its kernel?

Yes, it is possible for a linear transformation to have more than one subspace as its kernel. This can happen when there are multiple sets of vectors in the domain that get mapped to the zero vector in the range. In this case, all of these sets would be considered subspaces of the domain and would be kernels of the linear transformation.

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